Vertices of a triangle are mathrm{A}(-1,3), mathrm{B}(-2,2) and mathrm{C}(3,-1) . A new triangle is

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9.Straight Line

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The vertices of a triangle are $\mathrm{A}(-1,3), \mathrm{B}(-2,2)$ and $\mathrm{C}(3,-1)$. $A$ new triangle is formed by shifting the sides of the triangle by one unit inwards. Then the equation of the side of the new triangle nearest to origin is :

A

$x-y-(2+\sqrt{2})=0$

B

$-\mathrm{x}+\mathrm{y}-(2-\sqrt{2})=0$

C

$x+y-(2-\sqrt{2})=0$

D

$x+y+(2-\sqrt{2})=0$

(JEE MAIN-2024)

Solution

$Image$

equation of $\mathrm{AC} \rightarrow \mathrm{x}+\mathrm{y}=2$

equation of line parallel to $\mathrm{AC} \mathrm{x}+\mathrm{y}=\mathrm{d}$

$ \left|\frac{\mathrm{d}-2}{\sqrt{2}}\right|=1 $

$ \mathrm{~d}=2-\sqrt{2}$

$\mathrm{eq}^{\mathrm{n}}$ of new required line

$x+y=2-\sqrt{2}$

Standard 11

Mathematics

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